Show f(x) is a Lipschitz function

Let $(X,d)$ be a metric space and $a \in X$

Show that $f(x) = d(x,a)$ is a Lipschitz function from X to $\Re$

Use this fact to show $S$ is an open subset of $X$:

Let $(X,d)$ be a metric space and $a \in X, r > 0.$ Let $S$={$x \in X : d(a,x) > r$}

>Lipschitz Defn:

Let $f$ be a function from $(X,d)$ to $(Y,\rho)$

$f$ is said to be Lipschitz if:

$\exists$ K $\geq$ 0 s.t. $\rho( f(x_1) , f(x_2) ) \leq$ Kd($x_1$,$x_2$) $\forall$ $x_1 x_2 \in X$

>Continuity Defn:

Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and $f: X \to Y$

Then $f$ is continuous $iff$ f$^{-1}$($B$) = { $x \in X : f(x) \in B$} is an open subset of $X$, whenever $B$ is an open subset of $Y$

>Triangle Inequality for a Metric:

$d(x,z) \leq d(x,y) + d(y,z)$

My first question is, am I thinking about this correctly?

If $f(x)$ is a Lipschitz function, it is uniformly continuous

If $f(x)$ is uniformly continuous, it is continuous at every point

If $f(x)$ is continuous at every point $\Rightarrow$ the inverse images of open sets are open

Hence $S$ is an open subset of $X$.

Secondly, how would you formally construct this?

We know $f(x) = d(x,a)$ s.t. $f:X \to \Re$

By definition of Lipschitz we have $\exists K \geq 0$ s.t. $\rho(f(x_1),f(x_2)) \leq Kd(x_1,x_2) \forall x_1,x_2 \in X$

& since our codomain is $\Re$

$\rho(f(x_1),f(x_2)) = | d(x_1,a) - d(x_2,a) | \leq Kd(x_1,x_2)$

Taking $K=1$

$\ldots$

Help in finishing this off would be appreciated.

Thanks.

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Why did you use LaTeX only for part of the formatting? Please typeset all the math with LaTeX. – TMM Oct 31 '12 at 21:55

1 Answer

You can take $K=1$ in the definition.

For all $x_1,x_2\in X$, we have: $$d(x_1,a)\leq d(x_1,x_2)+d(x_2,a)$$ and $$d(x_2,a)\leq d(x_1,x_2)+d(x_1,a)$$ by the triangle inequality. We may rewrite these as $$d(x_1,a)-d(x_2,a)\leq d(x_1,x_2)$$ and $$d(x_2,a)-d(x_1,a)\leq d(x_1,x_2).$$ These two inequalities together mean precisely that $$|d(x_1,a)-d(x_2,a)|\leq d(x_1,x_2)$$ which is Lipschitz continuity of $f$ with constant $K=1$.

For the first question, you are correct. $S$ is open because $S=f^{-1}(r,\infty)$ and $f$ is continuous.

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I'm having trouble following this. I made a typo in the Lipschitz Defn above that I have corrected and I think you followed through with it. – user42538 Oct 31 '12 at 22:27
@user42538: I have reworded it a bit. I hope it's easier to follow now. In case something is not clear, you are welcome to ask. – Dejan Govc Oct 31 '12 at 22:36
Ah, completely my bad. Although I did make a typo, for some reason I was considering a as the other point, rather than looking at it as f(x_1) and f(x_2). Thanks. – user42538 Nov 1 '12 at 0:23